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olchik [2.2K]
3 years ago
8

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

is negligible.
(a) What is the kinetic energy of the ball as it leaves the hand?
(b) How much work is done by the gravitational force during the ball’s rise to its peak?
(c) What is the change in the gravitational potential energy of the ball during the rise to its peak?
(d) If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when it reaches the maximum height?
(e) What if the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand?
(f) What is the maximum height the ball reaches?
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

(a)0.0675  J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675  J

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

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ANTONII [103]

Answer:

The light bends away from the normal

Explanation:

We can solve the problem by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where:

n_1 is the index of refraction of the first medium

n_2 is the index of refraction of the second medium

\theta_1 is the angle of incidence (angle between the incoming ray and the normal to the interface)

\theta_2 is the angle of refraction (angle between the outcoming ray and the normal to the interface)

We can rearrange the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

In this problem, light travels from an optically denser medium to an optically rarer medium, so

n_1 > n_2

Therefore, the term \frac{n_1}{n_2} is greater than 1, so

sin \theta_2 > sin \theta_1\\\rightarrow \theta_2 > \theta_1

which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.

4 0
3 years ago
4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with
serious [3.7K]

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is \left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}

Substituting values

    3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s

Speed of block A after collision = 10 m/s

Option A is the correct answer.

5 0
3 years ago
Multiple choice 1. which of earth's spheres contains mountains, valleys, and other landscapes? (1 point
ycow [4]
Here are the answers:
1. Geosphere (though the term lithosphere is mostly used)
2. Both ice and wind (glaciers, and really strong winds)
3. Water
4. Its inertia (the Earth is constantly "falling" towards the Sun due to its gravitational pull, but its inertia helps the Earth from maintaining its orbit.)
5. The rotating Earth
6. one year
7. The equator
8. It depends on how much of the sunlit side of the Moon faces the Earth
9. When an object in space comes between the Sun and a third object
10. D<span>ifferences in how much the Moon and the Sun pull on different parts of Earth 
11. b. False
12. a. True

Hope my answers have come to your help.</span>
4 0
3 years ago
Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many
Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

8 0
3 years ago
An object moves in uniform circular motion at 25 m/s and takes 1.0 second to go a quarter circle. What is the radius of the circ
FromTheMoon [43]
Object Motion: 25 m/s

Circumference of Circle: 
1/4 Circumference of Circle in 1 second = 25 meters
25 meters times 4 = Circumference of Circle 
Circumference = 100 meters

Formula to Find Circumference of Circle: (work opposite)
C = 2<span>πr

100 = </span>2πr          divided
100/2π = r         simplify
50/π = r         (exact radius)

Answer:
50/π meters = r         (exact radius)

4 0
3 years ago
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